MS_Kekas wrote: For positive integers $a, b, c$ with sum $\frac{3}{2}$ How?

NO_SQUARES wrote: MS_Kekas wrote: For positive integers $a, b, c$ with sum $\frac{3}{2}$ How? Corrected, sorry

$a+b+c=\frac{3}{2} \to abc \leq \frac{1}{8}$ $\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} + \frac{1}{abc}=\frac{a^4+b^4+c^4}{abc}+\frac{1}{abc} \geq \frac{(a+b+c)abc}{abc}+\frac{1}{abc} \geq \frac{3}{2}+8=\frac{19}{2}$ And maximum not exists, because we can take $a=\epsilon, b=1,c=\frac{1}{2}-\epsilon$

For positive real numbers $a, b, c$ with sum $\frac{3}{2}$, prove that $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq\frac{3}{2}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} + \frac{1}{abc}\geq\frac{19}{2}$$Equality holds when $a=b=c=\frac{1}{2}.$

For positive real numbers $a, b, c$ with sum $\frac{3}{2}$, prove that $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ 55abc\geq\frac{67}{8}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ \frac{557}{10}abc\geq\frac{677}{80}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ \frac{5579}{100}abc\geq\frac{6779}{800}$$

sqing wrote: For positive real numbers $a, b, c$ with sum $\frac{3}{2}$, prove that $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ 55abc\geq\frac{67}{8}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ \frac{557}{10}abc\geq\frac{677}{80}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ \frac{5579}{100}abc\geq\frac{6779}{800}$$ Max $k$ such that for positive real numbers $a, b, c$ with sum $\frac{3}{2}$ true $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ (8k-12)abc\geq k$$ is $k_{max}=8.47378666059851...$ — root of polinom $16x^4-6156x^3+68688x^2-162837x+110889$

DipoleOfMonorak wrote: Max $k$ such that for positive real numbers $a, b, c$ with sum $\frac{3}{2}$ true $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+ (8k-12)abc\geq k$$ is $k_{max}=8.47378666059851...$ — root of polinom $16x^4-6156x^3+68688x^2-162837x+110889$ Let $c=\min(a,b,c)$ Rewrite the inequality as $F(a,b,c)=\frac{3\left(c^6+2(b+a)c^5+(b+a)^2c^4+((a^2-b^2)^2-25a^2b^2)c^2+2(b+a)(b^4+a^4)c+(b+a)^2(b^4+a^4)\right)}{abc((a-b)^2(7c+a+b) + (b-c)^2(7a+b+c) + (c-a)^2(7b+c+a))}\geq k$ then $k_{max}=\min(F)$ $F(a,b,c)-F\left(\frac{a+b}{2},\frac{a+b}{2},c \right)=\frac{3(a-b)^2(c+b+a)^2\cdot h(a,b,c)}{abc(a+b)^2(c+4a+4b)\left((a-b)^2(7c+a+b) + (b-c)^2(7a+b+c) + (c-a)^2(7b+c+a)\right)}$ where $h(a,b,c)=c^5+4(a+b)c^4-27abc^3-27ab(b+a)c^2+(b+a)^2(b^2-12ab+a^2)c+2(b+a)^3(2b^2+3ab+2a^2)$ $\Rightarrow~h(c+u, c+v, c)=$ $=96(v+u)c^4+(211v^2+269uv+211u^2)c^3+(v+u)(140v^2+173uv+140u^2)c^2+13(v+u)^2(3v^2+4uv+3u^2)c+2(v+u)^3(2v^2+3uv+2u^2)$ $\Rightarrow~h(a,b,c)\geq0$ for $c=\min(a,b,c)~\Rightarrow~F(a,b,c)\geq F\left(\frac{a+b}{2},\frac{a+b}{2},c \right)$ for $c=\min(a,b,c)$ $\Rightarrow~\min\left(F(a,b,c)\right)=\min\limits_{x\geq1}\left(F(x,x,1)\right)=\min\limits_{x\geq1}\left(\frac{3(8x^4+24x^3+15x^2+6x+1)}{2x^2(8x+1)}\right)=8.47378666059851...$

sqing wrote: For positive real numbers $a, b, c$ with sum $\frac{3}{2}$, prove that $$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq\frac{3}{2}$$$$\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} + \frac{1}{abc}\geq\frac{19}{2}$$Equality holds when $a=b=c=\frac{1}{2}.$ Solution. By Hölder’s Inequality, we have $$(b+c+a)(c+a+b)\left(\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \right)\geq(a+b+c)^3,$$which implies $\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \ge a+b+c= \frac{3}{2}$. In addition, the AM-GM gives \begin{align*}3\sqrt[3]{abc}\le a+b+c=\frac{3}{2}\implies\frac{1}{abc}\ge8. \end{align*}Therefore, $\frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab}+\frac{1}{abc}\ge\frac{3}{2}+8=\frac{19}{2}$, where the equality can occur for $a=b=c=\frac{1}{2}$. $\blacksquare$